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3 years ago

Can anyone please help me will mark brainliest

Mathematics

1 answer:

Andrei [34K]3 years ago

6 0

Answer: the price is $19.60

Step-by-step explanation:

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Ben’s garden supply store sells a gardening kit that normally costs $95. This week it is on sale for 35% off. How much does it c

gizmo_the_mogwai [7]

The gardening kit costs $33.25. Multiply $95 by 0.35.

3 0

3 years ago

Apply the distributive property to factor out the greatest common factor. 5 5 + 3 5 = 55+35=

vodomira [7]

Answer:

5(11+7) = 55+35

Step-by-step explanation:

The distributive property states that when multiplying a term with a sum multiply into each term within the sum. Here you have a sum. You’ll reverse the distributive property by factoring our a term which could multiply into each. 55 and 35 both have 5 as a factor. This makes the expression 5(11+7) = 55+35.

3 0

4 years ago

What are the roots of the polynomial equation x^3-5x+5=2x^2-5? Use a graphing calculator and a system of equations. Round nonint

leonid [27]

The right answer is c. –2.24, 2, 2.24

This question needs to be solved in two ways. First, using a graphing calculator. Next, using a system of equations.

1. Using a graphing calculator.

We have the following polynomial equation:

$x^3-5x+5=2x^2-5$

By ordering this equation we have:

$x^3-2x^2-5x+10=0$

So, we can say that this equation comes from a function given by:

$f(x)=x^3-2x^2-5x+10$

Thus, by plotting this function, we have that the graph of this function is indicated in Figure 1. By zooming, we can see, in Figure 2, that the roots of the polynomial equation are the x-intercepts of $f(x)$ which are:

$x_{1}=-2.236 \\ \\ x_{2}=2 \\ \\ x_{3}=2.236$

Finally, rounding noninteger roots to the nearest hundredth we have:

$\boxed{Root_{1}=-2.24} \\ \\ \boxed{Root_{2}=2} \\ \\ \boxed{Root_{3}=2.24}$

2. Using a system of equations.

The ordered equation is:

$x^3-2x^2-5x+10=0$

By arranging to factor out we have:

$x^3-5x-2x^2+10=0$

Then, by factoring:

$x(x^2-5)-2(x^2-5)=0$

Term $(x^2-5)$ is a common factor, thus:

$(x-2)(x^2-5)=0 \\ \\ (x-2)(x-\sqrt{5})(x+\sqrt{5})=0 \\ \\ Finally: \\ \\ \boxed{Root_{1}=-\sqrt{5}=-2.24} \\ \\ \boxed{Root_{2}=2} \\ \\ \boxed{Root_{3}=\sqrt{5}=2.24}$